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Series and Parallel resistors by Noora Aluami

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Series and Parallel resistors by Noora Aluami Powered By Docstoc
					 Series and Parallel resistors


By Noora Alnaimi
Lecturer: Dr. Rashid Alammari
Electrical Eng. Dept.
University of Qatar

                     Noora Alnuami   1
Headlines
 Definitions.
 Examples
 Practice
 Applications




                 Noora Alnuami   2
Definitions: Series Resistors
   Two resistors are in series if they
    exclusively share a single node and
    consequently carry the same current.




                     Noora Alnuami         3
Series resistors




            Those resistors are in series

                Noora Alnuami               4
 In the previous slide we can see that R1
  and R2 are in series so they will carry the
  same voltages and the share one single
  node..
 We can calculate R equivalent by the
  formula Req =R1+R2
 For N resistors in series :
                                     N
         Req=R1+R2+……………+RN =       ∑Rn   (1)
                                    n=1


                    Noora Alnuami               5
Definitions: Parallel Resistors
   Two resistors are in parallel if they
    connected to the same two nodes and
    consequently have the same voltage
    across them.




                     Noora Alnuami          6
Parallel resistors




                Noora Alnuami   7
 Those resistors are in parallel and they
  share two common node also they have
  they same voltage across them.
 If we want to calculate N number of
  resistors in parallel resistors we can do
  that by this formula..
1/Req=1/R1+1/R2+…….+1/RN

                    Noora Alnuami             8
For the circuit we can calculate R
  equivalent by another formula given the
  same result:
1/Req=1/R1+1/R2+….+1/RN
**If R1=R2=R3= …. .=RN then :
Req=R/N


                   Noora Alnuami            9
Example 1:
   Find Req in the circuit:




                      Noora Alnuami   10
 Solution:

From the left : we can see that 6 // 3  and
  we apply the formula 2..
Req=6*3/(6+3)=18/9= 2  so we will get
  ride of one resistor .
Then 1  is series with 5  so we can apply
  formula 1 :
Req=1+5=6 ..

                   Noora Alnuami            11
Those are in series




             Noora Alnuami   12
   Then 2  series with 2  so
    2+2=4 
   Then : 4( 2+2) // 6 so it gives:
    4*6/4+6=2.4 
   Then the all resistors are in series
    so we can apply formula (1)= R=4
     +2.4  +8  =
   14.4 




                                     Noora Alnuami   13
Example 2:




             Noora Alnuami   14
Solution:

   R3 and R 5 are not
    in parallel because
    there is a resistor
    between them.
   R1 and R2 are in
    parallel so
    Req=R1*R2/(R1+R
    2)=3/4 .


                      Noora Alnuami   15
  Then 3/4  is series with R3(2 )=
     3/4 +2=2.75 
 Then that result 2.75 is series with R4 =2.75  +10 =
     12.75 
 That result is parallel with R5 =
( 12.75 *2 ) /(12 .75+2  )=1.72 


   Then the 1.72  is series with R6 (12  ) : Req=1.72 +12
    =13.72 



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Practice : Find Req in the circuit     (Solution)




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R1 series with R2         RA =R1+R2
R5 series with RA         RB =R5+RA
R8 // RB                  RC =R8*RB/R8+RB
RC series with R4         RD = RC+R4
RD// R7                   RE =RD*R7/RD+R7
RE series with R3         RF=RE+R3
RF // R6                  RG=RF*R6/RF+R6
RG series with RA         Req=RG+RA.


                    Noora Alnuami           18

				
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